On differential subordinations for a class of analytic functions defined by a linear operator

(1)

PII. S0161171204309208 http://ijmms.hindawi.com © Hindawi Publishing Corp.

ON DIFFERENTIAL SUBORDINATIONS FOR A CLASS

OF ANALYTIC FUNCTIONS DEFINED

BY A LINEAR OPERATOR

V. RAVICHANDRAN, HERB SILVERMAN, S. SIVAPRASAD KUMAR, and K. G. SUBRAMANIAN

Received 25 September 2003

We obtain several results concerning the diﬀerential subordination between analytic func-tions and a linear operator deﬁned for a certain family of analytic funcfunc-tions which are in-troduced here by means of these linear operators. Also, some special cases are considered.

2000 Mathematics Subject Classiﬁcation: 30C45, 30C80.

1. Introduction. Let Ꮽ0 be the class of normalized analytic functions f (z) with f (0)=0 andf(0)=1 which are deﬁned in the unit disk∆:= {z∈C:|z|<1}. Let Ꮽbe the class of all analytic functionsp(z)withp(0)=1 which are deﬁned on∆. The classᏼofCarathéodoryfunctions consists of functionsp(z)∈Ꮽhaving positive real part. For two functionsf (z)andg(z)given by

f (z)=z+

∞

n=2

anzn_, _g(z)₌_z₊

∞

n=2

bnzn_, _(1.1)

their Hadamard product (or convolution) is deﬁned, as usual, by

(f∗g)(z):=z+

∞

n=2

anbnzn₌_:_(g_∗_{f )(z).} _(1.2)

Deﬁne the functionφ(a, c;z)by

φ(a, c;z):=

∞

n=0

(a)n (c)nz

n+1 _(c_≠₀_,₋₁_,₋₂_{, . . .}_; _z_∈_∆_), _(1.3)

where(x)nis the Pochhammer symbol or the shifted factorial deﬁned by

(x)n=

  

1, n=0,

x(x+1)(x+2)···(x+n−1), n∈N:= {1,2,3, . . .}. (1.4)

Corresponding to the functionφ(a, c;z), Carlson and Shaﬀer [1] introduced a linear operatorL(a, c)onᏭ0by the following convolution:

(2)

or, equivalently, by

L(a, c)f (z):=z+

∞

n=1

(a)n (c)nan+1z

n+1 _(z_∈_∆_). _(1.6)

It follows from (1.6) that

zL(a, c)f (z)=aL(a+1, c)f (z)−(a−1)L(a, c)f (z). (1.7)

For two functionsfandganalytic in∆, we say that the functionf (z)issubordinate tog(z)in∆, and write

f≺g or f (z)≺g(z) (z∈∆), (1.8)

if there exists a Schwarz functionw(z), analytic in∆with

w(0)=0, w(z)<1 (z∈∆), (1.9)

such that

f (z)=gw(z) (z∈∆). (1.10)

In particular, if the functiongisunivalentin∆, the above subordination is equivalent to

f (0)=g(0), f (∆)⊂g(∆). (1.11)

Over the past few decades, several authors have obtained criteria for univalence and starlikeness depending on bounds of the functionalszf(z)/f (z)and 1+zf(z)/f(z). See [4,5,7] and the references in [7]. In [2,6], certain results involving linear operators were considered. In this paper, we obtain suﬃcient conditions involving

L(a+1, c)f (z) L(a, c)f (z) ,

L(a+2, c)f (z)

L(a+1, c)f (z) (1.12)

for functions to satisfy the subordination

L(a, c)f (z)

L(a+1, c)f (z)≺q(z),

_L(a₊₁_{, c)f (z)}

L(a, c)f (z) β

≺q(z) q(z)∈Ꮽ. (1.13)

Also, we obtain suﬃcient conditions involving

L(a+2, c)f (z) L(a+1, c)f (z),

L(a+1, c)f (z)

z (1.14)

for functions to satisfy the subordination

_{L(a, c)f (z)}

z β

≺q(z), z

L(a+1, c)f (z)≺q(z)

(3)

SinceL(n+1,1)f (z)=Dn_{f (z)}_{, where}_Dn_{f (z)}_{is the Ruscheweyh derivative of}_{f (z)}_, our results can be specialized to the Ruscheweyh derivative and we omit these details. Note that the Ruscheweyh derivative of orderδis deﬁned by

Dδ_{f (z)}_:₌ z

(1−z)δ+1∗f (z)

f∈Ꮽ0; δ >−1 (1.16)

or, equivalently, by

Dδf (z):=z+

∞

k=2

δ+k−1 k+1

akzk f∈Ꮽ0; δ >−1. (1.17)

In our present investigation, we need the following result of Miller and Mocanu [3] to prove our main results.

Theorem1.1(cf. [3, Theorem 3.4h, page 132]). Letq(z)be univalent in the unit disk

and letϑandϕbe analytic in a domainD⊃q()withϕ(w)≠0, whenw∈q(). Set

Q(z):=zq(z)ϕq(z), h(z):=ϑq(z)+Q(z). (1.18)

Suppose that

(1) Qis starlike univalent in∆; (2) (zh(z)/Q(z)) >0forz∈ .

Ifp(z)is analytic in∆, withp(0)=q(0), p(∆)⊂D, and

ϑp(z)+zp(z)ϕp(z)≺ϑq(z)+zq(z)ϕq(z), (1.19)

thenp(z)≺q(z)andq(z)is the best dominant.

2. Main results. We begin with the following.

Theorem_2.1. _Let_α_,_β_{, and}_γ_{be real numbers,}_β≠_{0, and}₍₁+_{a)βγ <}_{0. Let}_q(z)∈Ꮽ be univalent inand let it satisfy the following condition forz∈ :

1+zq_q_(z)(z) >         

β+(1+a)γ

β if

β+γ(a+1)

β ≥0,

0 if β+γ(a+1)

β ≤0.

(2.1)

Iff (z)∈Ꮽ0and

L(a, c)f (z) L(a+1, c)f (z)

αL(a+1, c)f (z) L(a, c)f (z) +β

L(a+2, c)f (z) L(a+1, c)f (z)+γ

≺_a₊1₁α(a+1)+aβ+β+γ(a+1)q(z)−βzq(z),

(4)

then

L(a, c)f (z)

L(a+1, c)f (z)≺q(z) (2.3)

andq(z)is the best dominant.

Proof. Deﬁne the functionp(z)by

p(z):=_L(aL(a, c)f (z)₊₁_{, c)f (z)}. (2.4)

Then, clearly,p(z)is analytic in∆. Also, by a simple computation, we ﬁnd from (2.4) that

zp(z) p(z) =

zL(a, c)f (z) L(a, c)f (z) −

zL(a+1, c)f (z)

L(a+1, c)f (z) . (2.5)

By making use of the familiar identity (1.7) in (2.5), we get

L(a+2, c)f (z) L(a+1, c)f (z)=

1 a+1

1+ a

p(z)− zp(z)

p(z) . (2.6)

By using (2.4) and (2.6), we obtain

αL(a+1, c)f (z) L(a, c)f (z) +β

L(a+2, c)f (z) L(a+1, c)f (z)+γ

_{L(a, c)f (z)}

L(a+1, c)f (z)

=

_α

p(z)+ β a+1

1+_p(z)a −zp_p(z)(z) +γ

p(z)

=_a₊1₁(a+1)α+aβ+β+γ(a+1)p(z)−βzp(z).

(2.7)

In view of (2.7), the subordination (2.2) becomes

β+γ(a+1)p(z)−βzp(z)≺β+γ(a+1)q(z)−βzq(z) (2.8)

and this can be written as (1.19), where

ϑ(w):=β+γ(a+1)w, ϕ(w):= −β. (2.9)

Note thatϑ(w),ϕ(w)are analytic inC. Sinceβ≠0, we haveϕ(w)≠0. Let the functions Q(z)andh(z)be deﬁned by

Q(z):=zq(z)ϕq(z)= −βzq(z),

h(z):=ϑq(z)+Q(z)=β+(a+1)γq(z)−βzq(z). (2.10)

In light of hypothesis (2.1) stated inTheorem 2.1, we see thatQ(z)is starlike and _zh_(z)

Q(z)

= γ(a+1)+β −β +

1+zq

_(z)

q_(z) >0. (2.11)

(5)

Note that

L(1,1)f (z)=f (z),

L(2,1)f (z)=zf(z),

L(3,1)f (z)=zf(z)+z2f₂(z).

(2.12)

By takinga=c=1 inTheorem 2.1and after a change in the parameters, we have the following.

Corollary_2.2. _Let_α_{be a real number,}₁+_{α >}_{0, and let}_q(z)_{be univalent in}∆_, and let it satisfy

1+zq_q_(z)(z) >   −

α ifα≤0,

0 ifα≥0. (2.13)

Iff∈A0and

f (z) zf_(z)

(1−α)zf(z) f (z) −

1+zf_f_(z)(z) +α

≺zq(z)+αq(z)−α, (2.14)

then

f (z)

zf_(z)≺q(z) (2.15)

If we take

q(z)=1+₁₊λ_αz (2.16)

inCorollary 2.2, we obtain a recent result of Singh [7, Theorem 1(i), page 571]. By usingTheorem 1.1, we can show the following.

Lemma_2.3. _Let_γ_,_β_{be real numbers,}_β≠_{0, and}₁_{> γ/β}_{. Let}_q(z)∈Ꮽ_{be univalent} in∆and let it satisfy

1+zq_q_(z)(z) >          γ β if

γ β≥0,

0 if γ β≤0.

(2.17)

Ifp(z)∈Ꮽsatisﬁes

γp(z)−βzp(z)≺γq(z)−βzq(z), (2.18)

thenp(z)≺q(z)andq(z)is the best dominant.

(6)

Corollary2.4. Letα,β,γ be real numbers,β≠0, and1> γ/β. Letq(z)∈Ꮽbe univalent in∆and let it satisfy (2.17). Iff (z)∈Ꮽ0satisﬁes

f (z) zf_(z)

αzf

_(z)

f (z) +β

1+zf_f_(z)(z) +γ

≺α+β−βzq(z)+γq(z), (2.19)

then (2.15) holds andq(z)is the best dominant.

By usingTheorem 1.1, we obtain the following.

Theorem 2.5. Let a≠ −1. Let α, β, γ, and δ be real numbers, α≠0, and 1+

δ(a+1)(α+γ)/α >0. Letq(z)∈Ꮽbe univalent inand let it satisfy the following condition forz∈ :

1+zq(z)

q(z) >         

−δ(a+1_α)(α+γ) if δ(a+1)(α+γ)

α ≤0,

0 if δ(a+1)(α+γ)

α ≥0.

(2.20)

Iff (z)∈Ꮽ0and

αL(a+2, c)f (z) L(a+1, c)f (z)+β

_z

L(a+1, c)f (z) δ

+γ

_L(a₊₁_{, c)f (z)} z

δ

≺ α

δ(a+1)zq

_(z)₊_(α₊_γ)q(z)₊_β,

(2.21)

then

_L(a₊₁_{, c)f (z)}

z

δ

≺q(z) (2.22)

p(z):=

_L(a₊₁_{, c)f (z)}

z

δ

. (2.23)

Then, clearly,p(z)is analytic in∆. Also, by a simple computation, we ﬁnd from (2.23) that

zp(z) p(z) =

δzL(a+1, c)f (z)

L(a+1, c)f (z) −δ. (2.24)

By making use of the familiar identity (1.7) in (2.24), we get

L(a+2, c)f (z) L(a+1, c)f (z)=

1 δ(a+1)

zp(z)

(7)

By using (2.23) and (2.25), we obtain

αL(a+2, c)f (z) L(a+1, c)f (z)+β

_z

L(a+1, c)f (z) δ

+γ

_L(a₊₁_{, c)f (z)} z

δ

=α

₁

δ(a+1) zp(z)

p(z) +1 + β p(z)+γ

p(z)

=_δ(aα₊₁₎zp(z)+(α+γ)p(z)+β.

(2.26)

δ(a+1)(α+γ)p(z)+αzp(z)≺δ(a+1)(α+γ)q(z)+αzq(z) (2.27)

ϑ(w):=δ(a+1)(α+γ)w, ϕ(w):=α. (2.28)

Note thatϕ(w)≠0 andϑ(w),ϕ(w)are analytic inC. Let the functionsQ(z)andh(z) be deﬁned by

Q(z):=zq(z)ϕq(z)=αzq(z),

h(z):=ϑq(z)+Q(z)=δ(a+1)(α+γ)q(z)+αzq(z). (2.29)

By hypothesis (2.20) stated inTheorem 2.5, we see thatQ(z)is starlike and

_zh_(z)

Q(z)

= δ(a+1)(α+γ)

α +

1+zq

_(z)

q_(z) >0. (2.30)

Thus, by an application of Theorem 1.1, the proof of Theorem 2.5 is completed.

By takinga=c=1 inTheorem 2.5and after a suitable change in the parameters, we have the following.

Corollary_2.6. _Let_{α, β}≠₀_{be real and}₁+_{α >}_{0. Let}_q(z)_{be univalent in}∆_{and let} it satisfy (2.13). Iff∈A0and

βzf

_(z)

f_(z) +α

1−f(z)−βf(z)β≺zq(z)+αq(z)−α, (2.31)

then

f(z)β≺q(z) (2.32)

(8)

If we take (2.16) inCorollary 2.6, we obtain a recent result of Singh [7, Theorem 1(ii), page 571].

Theorem 2.7. Leta≠−1. Let α,β, and γ be real numbers and let β, γ ≠0and 1+α/γ >0. Letq(z)∈Ꮽbe univalent inand let it satisfy the following condition for z∈ :

1+zq(z)

q(z) >         

−α_γ if α γ ≤0,

0 if α

γ ≥0.

(2.33)

Iff (z)∈Ꮽ0and

_L(a₊₁_{, c)f (z)}

L(a, c)f (z) β

βγ

(a+1)L(a+2, c)f (z) L(a+1, c)f (z)−a

L(a+1, c)f (z) L(a, c)f (z) −1

+α

≺γzq(z)+αq(z),

(2.34)

then

_L(a₊₁_{, c)f (z)}

L(a, c)f (z) β

≺q(z) (2.35)

p(z):=

_L(a₊₁_{, c)f (z)}

L(a, c)f (z) β

. (2.36)

Then, clearly,p(z)is analytic in∆. Also, by a simple computation together with the use of the familiar identity (1.7), we ﬁnd from (2.36) that

1 β

zp(z)

p(z) =(a+1)

L(a+2, c)f (z) L(a+1, c)f (z)−a

L(a+1, c)f (z)

L(a, c)f (z) −1. (2.37)

Therefore, it follows from (2.36) and (2.37) that

_L(a₊₁_{, c)f (z)}

L(a, c)f (z) β

βγ

(a+1)L(a+2, c)f (z) L(a+1, c)f (z)−a

L(a+1, c)f (z) L(a, c)f (z) −1

+α

=γzp(z)+αp(z).

(2.38)

γzp(z)+αp(z)≺γzq(z)+αq(z) (2.39)

(9)

Note thatϕ(w)≠0 andϑ(w),ϕ(w)are analytic inC. Let the functionsQ(z)andh(z) be deﬁned by

Q(z):=zq(z)ϕq(z)=γzq(z),

h(z):=ϑq(z)+Q(z)=αq(z)+γzq(z). (2.41)

In light of hypothesis (2.33) stated inTheorem 2.7, we see thatQ(z)is starlike and

_zh_(z)

Q(z)

=

_α

γ+

1+zq_q_(z)(z) >0. (2.42)

Sinceϑandϕsatisfy the conditions ofTheorem 1.1, the result follows by an application ofTheorem 1.1.

By takinga=c=1 inTheorem 2.7and after a suitable change in the parameters, we have the following.

Corollary2.8. Letα, β≠0andγbe real with1+α/γ >0. Letq(z)be univalent in ∆and let it satisfy (2.33).

Iff∈A0and

_zf_(z)

f (z) β

βγ

1+zf_f_(z)(z)−zf_{f (z)}(z)

+α

≺γzq(z)+αq(z), (2.43)

then

_zf_(z)

f (z) β

≺q(z) (2.44)

If we take (2.16) andγ=1 inCorollary 2.8, we obtain a recent result of Singh [7, Theorem 1(iii), page 571] and, by setting

q(z)=

1

0

1−λztα

1+λztαdt (2.45)

andα=1 inCorollary 2.8, we obtain another recent result of Singh [7, Theorem 3, page 573].

Theorem 2.9. Letα≠0 andγ be real numbers,(a+1)αγ <0. Letq(z)∈Ꮽ be univalent inand let it satisfy the following condition forz∈ :

1+zq(z)

q(z) >         

α+γ(a+1)

α if

α+γ(a+1)

α ≥0,

0 if α+γ(a+1)

α ≤0.

(10)

Iff (z)∈Ꮽ0and

α L(a, c)f (z) L(a+1, c)f (z)

_L(a₊₂_{, c)f (z)}

L(a+1, c)f (z) +γ

L(a, c)f (z) L(a+1, c)f (z)

≺_aaα₊₁+

_α

a+1+γ q(z)− α a+1zq

_(z),

(2.47)

then

L(a, c)f (z)

L(a+1, c)f (z)≺q(z) (2.48)

The proof of this theorem is similar to that ofTheorem 2.1and hence it is omitted. By takinga=c=1 inTheorem 2.9and after a suitable change in the parameters, we have the following.

Corollary_2.10. _Let₀≤_α≤₁_and_q(z)_{be univalent in}∆_{and let them satisfy}

1+zq_q_(z)(z) >   

α ifα≥0

0 ifα≤0. (2.49)

Iff∈A0and

f (z) zf_(z)

1+zf_f_(z)(z) −α _{f (z)}

zf_(z)−1 ≺(1+α)−αq(z)−zq(z), (2.50)

then (2.15) holds andq(z)is the best dominant.

Let

q(z)=1+λz k

1

0

tα

1+(z/k)tdt. (2.51)

After a change of variable in (2.51), we get

q(z)=1+_zλα z

0

ηα

k+ηdη. (2.52)

By diﬀerentiating (2.52), we have

zq(z)= λz

k+z−αq(z)+α (2.53)

or

α−αq(z)−zq(z)= −_kλz₊_z. (2.54)

Since the bilinear transform

w= − λz

(11)

maps∆onto the disk

w+₁₋λ_k₂≤_k|₂λ₋|k₁, (2.56)

fromCorollary 2.10for the functionq(z)given by (2.51), we obtain a recent result of Singh [7, Theorem 2(i), page 572].

Theorem_2.11. _Let_α≠₀_and_γ _{be real numbers,}_(a+₁_{)αγ <}_{0. Let}_q(z)∈Ꮽ _be univalent inand let it satisfy (2.46) forz∈ .

Iff (z)∈Ꮽ0and

αzL(a+2, c)f (z) L(a+1, c)f (z)2+γ

z

L(a+1, c)f (z)≺(α+γ)q(z)− α a+1zq

_(z), _(2.57)

then

z

L(a+1, c)f (z)≺q(z) (2.58)

The proof of this theorem is similar to that ofTheorem 2.1and therefore it is omitted. By takinga=c=1 in Theorem2.11and after a suitable change in the parameters, we have the following.

Corollary2.12. Let0≤α≤1andq(z)be univalent in∆and let them satisfy (2.49). Iff∈A0,f (z)f(z)/z≠0, and

zf(z) f(z)2 −α

₁

f(z)−1 ≺α−αq(z)−zq

_(z), _(2.59)

then

1

f_(z)≺q(z) (2.60)

On setting (2.51) in Corollary 2.12, we obtain a recent result of Singh [7, Theorem 2(ii), page 572].

Acknowledgment. The authors are thankful to the referees for their comments.

References

[1] B. C. Carlson and D. B. Shaﬀer,Starlike and prestarlike hypergeometric functions, SIAM J. Math. Anal.15(1984), no. 4, 737–745.

[2] J.-L. Liu and S. Owa,On a class of multivalent functions involving certain linear operator, Indian J. Pure Appl. Math.33(2002), no. 11, 1713–1722.

[3] S. S. Miller and P. T. Mocanu,Diﬀerential Subordinations. Theory and Applications, Mono-graphs and Textbooks in Pure and Applied Mathematics, vol. 225, Marcel Dekker, New York, 2000.

(12)

[5] K. S. Padmanabhan,On suﬃcient conditions for starlikeness, Indian J. Pure Appl. Math.32

(2001), no. 4, 543–550.

[6] J. Patel and P. Sahoo,Properties of a class of analytic functions involving a linear operator, Demonstratio Math.35(2002), no. 3, 495–507.

[7] V. Singh,On some criteria for univalence and starlikeness, Indian J. Pure Appl. Math.34

(2003), no. 4, 569–577.

V. Ravichandran: Department of Computer Applications, Sri Venkateswara College of Engi-neering, Pennalur, Sriperumbudur 602 105, Tamil Nadu, India

E-mail address:[email protected]

Herb Silverman: Department of Mathematics, College of Charleston, Charleston, SC 29424, USA E-mail address:[email protected]

S. Sivaprasad Kumar: Department of Mathematics, Sindhi College, 123 P.H. Road, Numbal, Chennai 600 077, Tamil Nadu, India

E-mail address:[email protected]

K. G. Subramanian: Department of Mathematics, Madras Christian College, Tambaram, Chennai 600 059, Tamil Nadu, India